Fig. 8. B. 1, PROP. V. THEOR. If a right line (AB) be cut into equal parts (in C) and into unequal (in D), the rectangle under the unequal parts (AD and DB), together with the square of the intermediate part (CD) is equal to the square of the half line (CB). On CB describe the square CKMB, draw KB, and through the point D draw DL parallel to CK, and cutting KB in G and through G draw HGE parallel to AB, until it meet the line AE drawn through A parallel to CK. Because the lines AC and CB are equal (1), the (1) hypoth. (2) Prop.36 rectangles AF and CH are equal (2), but the rectangles CG and GM are also equal (3), therefore the rectangle AG is equal to the gnomon CHL (4), add (4) Ar. 2. to both the square FL, and the rectangle AG together (5) Cor. with the square FL is equal to the square CKMB, Prop. 45. But the rectangle AG is the rectangle under AD and B. 1.& def. DB, for DG is equal to DB (5), and FL is the square (6) Prop.34 of CD, because FG and CD are equal (6), and CKMB B. 1. is the square of CB. 31. B. 1. Fig. 9. (1) Prop. 1. B. 2. Otherwise thus: The rectangle under AD and DB is equal to the sum of the rectangles under AC and DB and under CD and DB (1), but the rectangle under AC and DB is equal to the rectangle under CB and DB, because AC and CB are equal, or to the rectangle under CD and DB, together with the square of DB (2), add to (2) Prop. 3. both the square of CD, and the rectangle under AD and DB together with the square of CD is equal to double the rectangle under CD and DB together with the squares of CD and DB, that is, to the square of CB (3) Prop 4. (3). B. 2. B. 2. Cor. 1. Hence it is evident that if any line be bisected, the rectangle under the parts is greater than if it be cut unequally, and therefore the sum of the squares of the parts is less (3). Cor. 2. If two equal right lines be so divided that the rectangle under the segments of one be equal to the rectangle under the segments of the other, their segments shall be equal. Fig. 10. (1)hypoth. If one of the right lines be bisected, then the other is also bisected, for the rectangle under the parts is then equal to the square of the half line, whence it is evident that the segments themselves are equal. But if they be not bisected, let AB and CD be the given equal lines, and let them be divided in E and F. Let the lines be bisected in G and H, and as the lines are equal(1), their halves are equal, and therefore the squares of the halves (2), but the rectangles under AE (2) Cor. 1. and EB, and under CF and FD are also equal(1), take p. 46. B.l. away these from the equal squares, and the remainders, that is, the squares of GE and HF (3) are equal, and (3) Prop.5. therefore the lines themselves GE and HF are equal(4), B. 2. therefore the sums and differences of them and the (4) Cor. 2. halves, that is, the segments AE and CF, EB and FD B. 1. are equal. Prop. 46. Cor. 3. The rectangle under the sum and difference Fig. 9. of two right lines is equal to the difference between their squares. Because that rectangle together with the square of the less is equal to the square of the greater, as is evident from the preceding proposition, for AC is the greater line, CD the less, and DB the difference, Schol. 1. Hence it is evident that in a right angled triangle, the rectangle under the sum and difference of the hypothenuse and one side is equal to the square of the other side. Schol. 2. Hence it is also evident that if from any angle of a triangle a perpendicular be drawn to the opposite side, the rectangle under the sum and difference of the sides is equal to the rectangle under the sum and difference of the segments of the side to which the perpendicular is drawn. For the rectangle under the sum and difference of the sides is equal to the difference between the squares of the sides, and the rectangle under the sum and difference of the segments of the base is equal to the dif ference between the squares of the segments, and these differences are equal by Cor. 4. prop. 47, book 1, Fig. 11. Cor. 4. The difference between the squares of two sides BA and AC of any triangle BAC, is equal to double the rectangle under the remaining side BC, and the distance from its middle point D of the perpendicular AF, which is drawn to it from the opposite angle. For the difference between the squares of the sides AB and AC is equal to the difference between the squares of the segments, BF and FC, of the side upon (1) Cor. 4. which the perpendicular falls (1), and, therefore, when Prop.47 B.1 the perpendicular falls within the triangle, to the rectangle under BC the sum of these segments, and (2) Cor. 3. the difference between BF and FC (2), or to double Prop.5.B.2 the rectangle under BC and half the difference, that is, the distance of the point F from the middle point of the side BC. But if the perpendicular fall without the triangle, the difference of the squares of the segments BF and FC is equal to the rectangle under BC, the difference, and the sum of BF and FC, or to double the rectangle under BC and half the sum of BF and FC, that is, the distance of the point F from the middle point of the side BC. Schol. Hence, given in numbers the sides of any triangle we can find its area; divide the difference between the squares of two sides of the triangle by the remaining side, to half this side add half the quote, and subtract the square of this sum from the square of the greater side, the remainder is the square of the perpendicular, thence the perpendicular itself is found, which multiplied into half the side upon which it falls gives the area of the triangle. PROP. VI. THEOR. If a right line (AB) be bisected (in C), and pro- Fig. 13. duced to any point (F), the rectangle under the whole see N. line thus produced (AF), and the produced part (BF) together with the square of the half (CB) is equal to the square of (CF) the line made up of the half and produced part. On CF describe the square CEGF, draw EF, and through the point B draw BP parallel to FG, and cutting EF in K, through K draw LO parallel to CF, and meeting AO which is drawn through A parallel to CD. B. 1. Because AC and CB are equal (1), the rectangle AD (1) hypoth is equal to the rectangle CK (2); but the rectangles CK (2) Prop.36 and KG are equal (3), therefore AD is equal to KG, add to both CL, and AL is equal to the gnomon CLP, B. 1. add to both DP, and the sum of AL and DP is equal (4) Cor. to the square of CF. But AL is the rectangle under Prop.43.B. the whole produced line and the produced part, for 1.&def. 31. FL is equal to BF (4), and DP is the square of the (5) Cor. balf CB, for it is the square of DK (5), and DK is Prop.43 B1 equal to CB (6). (6) Prop.34 B. 1. Otherwise thus: B. 1. B 2. The rectangle under AF and BF is equal to the sum of the rectangles under AC and BF, and under CB and BF, together with the square of BF (1), but (1)Prop. 1. the rectangle under AC and BF is equal to the rectangle under CB and BF, as AC and CB are equal, therefore the rectangle under AF and BF is equal to double the rectangle under CB and BF together with the square of BF, add to both the square of CB and the rectangle AFB, together with the square of CB, is equal to double the rectangle CBF, together with the squares of CB and BF, that is, to the square of CF (2). (2) Prop. 4. Cor. If a right line BE be drawn from the vertex of B. 2. an isosceles triangle to the base, the rectangle un G 4 der the segments of the base AE and EC is equal to the difference between the square of this line BE and the square of either side AB or BC. Bisect AC in D, and draw the right line BD. The rectangle AEC is equal to the difference be(1)Prop. 5. tween the squares of DC and of DE (1), add to both & 6. B. 2. the square of the perpendicular BD, and the rectangle AEC is equal to the difference between the sum of the squares of DC and DB or the square of BC, and the sum of the squares of DE and DB or the square of EB. But if the line drawn to the base be perpendicular to it, it bisects the base, and the rectangle under the segments of the base is the square of the half, and (2) Prop.47 therefore together with the square of the line drawn to the base, equal to the square of the side (2). B. 1. Fig. 15. (1) Cur. Prop. 43. B. 1. PROP. VII. THEOR. If a right line (AB) be divided into any two parts, the squares of the whole line (AB) and either segment (CB) are together equal to double the rectangle under the whole and that segment, together with the square of the other segment. Describe the square of AB, draw FB, through the point C draw CG parallel to AF, and through P, its intersection with FB, draw DE parallel to AB. The square AK is equal to the rectangles AE and PK, together with the square DG: add to both the square CE, and the squares AK and CE taken together, are equal to the rectangles AE and CK together with the square DG. But AE is equal to the rectangle under AB and CB, because CB and BE are equal (1), and CK is also (2) Def. 31. equal to the rectangle under AB and CB, because KB is equal to AB (2), and DG is the square of AC, because DP and AC are equal (3). B. 1. (3) Prop. 34, B. 1. |